Answer:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;
[tex]a_t = ar[/tex]
where;
a is the angular acceleration and
r is the radius of the circular path
[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]
Determine time of the rotation;
[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]
Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration
[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]
The magnitude of total linear acceleration is given by;
[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]
Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
